Sure, let's solve the given question step-by-step.
Given: [tex]\( 12 \cdot {}^4C_2 = {}^{2n}C_1 \)[/tex]
Recall the combination formula:
[tex]\[
{}^nC_r = \frac{n!}{r!(n-r)!}
\][/tex]
First, let's compute both combinations involved in the problem:
1. For [tex]\( {}^4C_2 \)[/tex]:
[tex]\[
{}^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1) \cdot (2 \cdot 1)} = \frac{24}{4} = 6
\][/tex]
2. For [tex]\( {}^{2n}C_1 \)[/tex]:
[tex]\[
{}^{2n}C_1 = \frac{(2n)!}{1!(2n-1)!} = \frac{2n}{1} = 2n
\][/tex]
Substitute these results into the given equation:
[tex]\[
12 \cdot 6 = 2n
\][/tex]
Simplify:
[tex]\[
72 = 2n
\][/tex]
Solve for [tex]\( n \)[/tex]:
[tex]\[
2n = 72 \implies n = \frac{72}{2} = 36
\][/tex]
Therefore, the value of [tex]\( n \)[/tex] is:
[tex]\[
\boxed{36}
\][/tex]
